First, we find the unit tangent vector $T=\frac{r'(t)}{|r'(t)|}$: $$r'(t)=\left\langle 1,-\frac{\sin t}{\cos t}\right\rangle$$ $$|r'(t)|=\sqrt{1+\frac{\sin^2t}{\cos^2t}}=\frac{1}{\cos t}$$ $$T=\frac{r'}{|r'|}=\cos t\left\langle 1,-\frac{\sin t}{\cos t}\right\rangle= \langle \cos t,-\sin t\rangle$$
Next, we find the curvature $\kappa(t)=\frac{|T'(t)|}{|r'(t)|}$:
$$T'(t)=\langle -\sin t,-\cos t\rangle$$ $$|T'(t)|=\sqrt{\sin^2t+\cos^2t}=1$$ $$\kappa(t)=\frac{1}{\frac{1}{\cos t}}=\cos t$$
However, when I tried to check in Mathematica, I got the following output:
ArcCurvature[{t, Log[Cos[t]]}, t] $$\frac{| \sec (t)| ^2}{\left(\tan ^2(t)+1\right)^{3/2}}$$
Be careful with mod signs. Their correct answer simplifies to $\sec^2t/|\sec^3t|=|\cos t|$. You should have said $|r^\prime|=|\sec t|$.